Best guess at the Pascal's Triangle: This goes back to what you were doing at the beginning, expressing each binomial expansion as the previous one multiplied by another (A+B). In that system, each of A and B gets multiplied by each of the previous parts, and the sum of all those results gives us our next expansion. So let's look at the third expansion: A^3+3(A^2)B+3A(B^2)+B^3. To find our fourth, each of those gets separately multiplied by an A and a B. To simplify, let's look at that with just the first two terms: (A^3+3(A^2)B)*(A+B). If we multiply A^3 by A, we get A^3. If we multiply it by B, we get (A^3)B. If we multiply 3(A^2)B by A, we get 3(A^3)B. If we multiply it by B, we get 3(A^2)(B^2). Now the question is, how many (A^3)B results did we get, total? Well, we got one for every A^3 in the previous expansion (1 total), plus one for every (A^2)B in the previous expansion (3 total). No other variable segment can produce (A^3)B, so our result is the sum of those two terms. Since each variable segment can only become one of two different variable segments in the next expansion, this should happen with any other pair of terms as well. (This one also gave us a bonus term because the A^4 term can only be approached by one segment, but that won't happen with more central segments.) That took a lot of work to figure out. I should do this stuff more often...
As soon as I saw the 4641 I though Pascal’s Triangle must be coming. I use the first 4 rows when I want an example for some compound interest in class and just use 10%.
I swear to god that if I'd had you as an AS Maths teacher instead of the disinterested person I did get, I wouldn't have miserably failed it haha. Another superb video, entertaining and well laid out :) Thanks, and I hope to keep seeing em coming!
So nice of you to say that my comment was your favourite =) Regarding Pascal triangle it is pretty easy to see why it works In binomial coefficient we have for example 1 3 3 1 Then when we multiply by (a + b) if we consider the result it will be 1 3 3 1 0 + 0 1 3 3 1 = 1 4 6 4 1 Because powers will be shifted by a or b, so we sum up previous coefficient with itself but shifted by 1 position. And pascal triangle does exactly that, you sum up (k,n) position with (k,n+1) position in order to get the position (k+1, n+1) (where k is the number of row in Pascal triangle), and positions of (k, 0) and (k,k) are always 1 (which even works for 0's row, because 1 = 1) P.S. Binomial Coefficient is one of my favourite subjects though, I was so interested in this concept that I actually asked my dad (who has PhD in Physics) to explain it to me when I was in 3rd Grade... Though I asked him to explain the Choose Function back then (even though I didn't know the word "function" back then), only later on I found out about Binomial Coefficients and Pascal Triangle. And I personally prefer "Binomial Coefficient" or "Newton's Binomial Series" names.
I noticed the Pascal's triangle thing when you wrote the numbers down, but I couldn't figure out why. I'm really looking forward to you explaining why :)
+Harm Prins Hint: You can already find the answer in this video if you look at the way terms are organized in the computation of (a+b)^4 from (a+b)^3 x (a+b).
Harm Prins the explanation that seems most likely to me is that each number on Pascal's triangle lists the number of ways to reach that location from the top of the triangle, moving downward through adjacent numbers. there's only one way to reach the top number: by going nowhere. there is only one way to reach the first number on the second row, which is by moving one down to the left there is only one way to reach the second number on the second row, which is by moving one down to the right. there are two ways to reach the second number on the third row. one is by moving right, then left. the other is by moving left, then right. and so forth. I'm sure there's a more rigorous proof behind that, but that seems to be the gist of it
Pascal's triangle answer: the number of paths you can take from the top to a number. i.e. you can take 6 different paths to get from the 1 at the top to that number 6. What a fun way to wake up on this post finals day. :D Thank you very much, Matt!!
It's clear from the definition of the choose function that for any natural number n, C(n, 0) = C(n, n) = 1. It is relatively straightforward to show - using the definition of the choose function and some algebra - that for any natural number m such that 0
At this point i have to: Zeroth: wander what the hell will happen if you make a triangle with trinomial coefficients. 1/2th: sugest it has to do with "Pascal's tetrahedron" First: admit you've blown my mind Second: thank you this awesome video Third: point out that at the time I'm writing this video has 234 likes and 2 dislikes
Since the moment I ended writing my comment I started thinking about the zeroth point, and I foun that not only I was right in the 1/2th point but this is true for any given dimension. I've tryed it for the first five dimensions and I'm amazed of how quickly they grow in size. I was also very glad about knowing that all the time I spent calculating and writing Pascal's n-simplex down in my window.
Love these videos so far, it's like having extra Numberphile videos to watch! I just wanted to ask if I may, do you have an intuitive understanding of maths in any way or is it something you've had to work extra hard on? I ask because while I love maths I really struggle with it, which is why I love videos like this. Makes it all that bit easier to grasp.
Another truly *standup video!* Splendid! You never sat down once! I was breathlessly waiting for you to show the direct connection between Choose and Binomial by showing the Choose function being used to 'assemble' each term in the expansion. And son-of-a-gun if you didn't do exactly that! Kudos, Matt, and thanks! Fred
When you are at 121 it means that you have 1 way to have aa 2 ways to have ab and 1 way to have bb. When you start the 1331 the first 1 is the number of ways to get aaa, which is the 1 way to get aa with an extra a. The first 3 is the number ways to get aab, and that is the 1 way to get aa with an extra b plus the 2 ways to get ab with an extra a. And that is - by example - why you can get the coefficients by adding pairs of the coefficients for the previous power.
Apart from the fact that I have this geniunely funny guy explaining Math to me, I'm also satisfied that he cares to explain "why". This is something noone does.
I don't care much about the maths, but I watched the whole damn video because I love your passion and enthusiasm. Would gladly watch more. Wish I had someone like you as a school teacher all those years ago.
For every number in the triangle, N is the number of ways to get back to the top, moving either left or right for every tier on the way up. For the 1s on the edges, they have no choice but to keep going on the edge. The first 2 can go either right-left or left-right. The first 3 can go left-right-right, right-left-right, or right-right-left. The first 4 can go left-right-right-right, right-left-right-right, right-right-left-right, or right-right-right-left. That's all boring because it's on the edges while there is some "choice" in the order of the rights and lefts, you still only choose one left. Hence, for the 4 it is 4 choose 1 or 4C1. For the 6, now, it can go R-R-L-L, R-L-R-L, R-L-L-R, or the mirror images of all of those, since it's in the middle. Here, we are picking 2 lefts, so it is 4C2. Then it's 4C1 again and finally 4C0. As to _why_ these numbers are the way to get back to the top, I'd wager it's because every extra layer is basically another "term" of the binomial where you ought to choose a or b, in this case right and left. As you add a layer, terms build up, and so the layer inherits the terms of the previous one. And adding the two numbers above is adding the two paths to get back to the top from each of those instances. The 6 has two ways up - the two 3s. Each of those 3s has 3 paths up, so 3 + 3 = 6. So there you go. The Pascal Triangle works because you add the two numbers of paths above it, since you need to choose one out of two of them. And that's nested, and as we know, nested things tend to make fractals, and oh dear it seems Pascal's triangle has a link to the Sierpinski triangle ;) And Matt, if you're reading this, hi!
+standupmaths Instead of explaining why it relates to the choose function directly, show that adding a row to the triangle is a skewed version of multiplying by 11. 11^n = (10 + 1)^n, so you have binary coefficients multiplied by powers of 10. You've already proven binary coefficients are related to the choose function, so the numbers in Pascal's triangle are too. QED.
+standupmaths Seriously. Come teach me, because my maths teacher is the worst. I love maths, but she ruins the lessons I have. Also, I have a test tomorrow. :(
Yaaaaay! Matt is the best. I absolutely positively love Pascal's Triangle, & it overjoyed me in my sophomore (10th grade) algebra class when I learned that you could find binomial coefficients with it. Thank you, Mr Parker! :) So Long & Thanks For All the Fish, Lawrence Calablaster
I'm a college student (year 13 UK) and I just want to say I love your videos and that I just did this in class Monday I found this by chance and it has really helped thanks for the great videos
The pascal's triangle works because you are multiplying the previous term including the corresponding variables by x and y. For example, multiplying x³y² by x gives you x⁴y² but multiplying it by y gives you x³y³. But you could also get x⁴y² by multiplying x⁴y by y, and you could also get x³y³ by multiplying x²y³ by x. This is why each number gets added twice, once to the number on the left, and once to the number on the right. This also explains why if you add up all the numbers in each row, it will give you increasing powers of two as you go down. Because each number is getting added twice, its the same as multiplying everything by two.
Video tips: the best way to fix the focus problem is (1) forget autofocus and use manual focus on the whiteboard and (2) increase the lighting so that your camera uses a smaller aperture. If you want better audio you could also wear something with a higher neckline and get the mic closer to your mouth, without obstruction from clothing... Great video though! I kept wondering why you hadn't mentioned Pascal's triangle... and then you did :)
Each position is the number of paths to that position from the beginning if every movement is a downward diagonal, like checkers. Which is essentially choosing a number of left and right moves, a and b moves, which always have to add up to the same amount to get to the same position, all the ways to get 3 Lefts and 4 Rights, which is the choose function. Which isn't the Why the numbers are that so much as the surface connection I'll probably come back and have a go at that later.
Explanation at 9:08 was initially very confusing for me because the situation seemed more like it required permutations (P) rather than choose (C). Perhaps it is useful to imagine the a or b values as being separate for the purposes of the explanation, which can be done through assigning a subscript value. Then the reason why the n value in n choose r equals the power of the binomial is because this power of the binomial is the same as the number of "separate" a or b values which can be picked. Other than that, thanks for the great video!
I'm from indian you are great explanation of this topic Keep it up ALWAYS You know ours mathematicians. Like RAMANUJAN. ARYABHATTA. SAKUNTALA DEVI All are genius You also like that Thanks to give your time to read my comment I salute you
I cant believe I watched this video when it only had 18 views! I'd like to say that's a good explanation, and thank you for spending your time making awesome videos!
Here it is! : Each number in Pascal's triangle is the ammount of paths you can take to get up. So why can you calculate a number by adding the two above it? because after you choose one of them (let's say the number's 84), there are 84 paths to continue after that, so if you add it up with the paths to the other option you get all the paths for that number. Starting at the top (it's easier to visualize), to get to any position in the triangle you need a very specific ammount of turns to the left and to the right, because anything else would leave you in another square; however, these turns can be in any order and you'll still get to the right position. So, let's start by analyzing the leftmost side: for all steps you need 0 right turns and S (the number of the step counting from the top) left turns. S choose 0 returns 1 for any S, so that's why all numbers in the left are 1s After you have that it's really easy to continue: the second squares from the left require one step to the right, so you do S choose 1 wich varies depending on S; the next square requires 2, the next 3 and so on... You can also count the number of left turns it needs and get the same result because of symmetry, or count either the right or left turns starting from the right. So from what i've explained above, we get this: If you're in the Nth step (from top to bottom), you calculate the Mth number (from left to right, anything from 0 to N) by doing N choose M, or choose(N,M) as i prefer to say it (because programming) Oh! also,adding up the numbers in a step gets you a power of 2 because you can express all L/R paths as binary numbers! and each step uses all possible paths.
As you go down the triangle, you can go either left or right. So, for the 1,4,6,4,1 line, of the 4 steps you go down, you would need to go left twice and right twice to get to 6, i.e 4C2
Also, if you cut parallel lines at a jaunty angle through Pascal's triangle, you get the Fibonacci sequence. 1=1 1=1 1+1=2 1+2=3 1+3+1=5 1+4+3=8 And so on
Pretty sure i got it! i'll write it later because i can't now but it relates to binary and how many ways there are to arrange a certain number of ones or zeros (using the choose function)
Just woke from a dream where i was trying to explain to someone the connection between pascal's triangle and factorising polynomial functions to solve trivial problems in Asymmetric cryptography and now ive woken up and here i am. Thanks for the reminder :)
I know there tons of numberphile videos about this, is 1 + 2 + 3 + 4 + 5... = -1/12? Or is there something wrong with this proof. I would like to see more videos about infinite sums. I love how you keep making more videos. I thought this channel was going to go dead but now there are more Math videos. YAY!
The -1/12 thing is a world of confusion but I might take it on one day. And yes: I'm trying to keep this channel alive! Hard to find the time+money though.
I have a TI-30XS MultiView calculator that has a "table" button, with which I can input an expression with one variable (x), and it returns a list of appropriate y values. I can't use the proper notation, but instead I have to use things like "n nCr x. This works well for the rows of Pascal's Triangle, but I found a way to generate the sequences of numbers in the diagonals of the Triangle: if I use x nCr n (for example, x nCr 2) I get the triangular numbers. If I use x nCr 3, I get the tetrahedral numbers, etc. This works for any of the diagonals of the Triangle! 🤓
Can you please please please make a video about trig identities? I wanna know why sin(a+b) is equal to sinx cosy + siny cos x and same to cos and tan addition. You and Numberphile are the only channels that makes math sound exciting. Thank you
Sadly school teachers have to cover the curriculum which leaves very little time for anything else like this. And if a teacher does make time, students/parents complain that it's not going to be on the test.
The pascal's triangle gives a count of the number of shortest paths from the top to the current number. Why does this happen: Each number has two "parent" numbers (or one in case of the outside numbers of the triangle). The only possible way to arrive at the current number (on a shortest path) is via one of these parents. If the parents number display the number of paths to their number, adding up these parents will result in the amount of paths to the current number. This reasoning can be from the bottom up to the second row of the triangle (at 1 1) . Because these 1's are the number of paths to these numbers (both 1's can only be reach in one way), the theory is correct. For all the numbers with only 1 parent, there is only 1 shortest path to reach these numbers and they therefore always contain a 1. In short each number has two parents with a count of the total number of paths to these parents, by summing them the total number of paths to the current number is obtained. Why does this give the correct result: If we take a look at the 5th row of the triangle one path two the 6 could be described by LRLR, in which the L and R give a choice to either the left or the right respectively. The path would in this way follow the numbers: 11236. To arrive at the 6, precisely two of the 4 choices must be Right (the other two automatically being a Left) or in other words you have 4 options and you have to choose 2 correct ones out of it, which brings you right back to the start of the video (the choose function/binomial coefficients).
Hi, Matt. Perhaps this will interest you. A variation on Pascal's triangle yields both the Fibonacci sequence from F[3] (2, 3, 5, 8, 13,...) and the Lucas numbers from L[1] (1, 3, 4, 7, 11,...), by summing the descending and ascending shallow diagonals respectively: 1 2 (Σ = 3 = 3 * 1) 1 3 2 (Σ = 6 = 3 * 2) 1 4 5 2 (Σ = 12 = 3 * 4) 1 5 9 7 2 (Σ = 24 = 3 * 8) 1 6 14 16 9 2 (Σ = 48 = 3 * 16) 1 7 20 30 25 11 2 (Σ = 96 = 3 * 32) 1 8 27 50 55 36 13 2 (Σ = 192 = 3 * 64) This triangle also yields: the natural numbers from 2; the odd numbers; the squares; the square pyramidal numbers; numbers of the form n(n + 3)/2 (similar to triangular numbers); the sequence S[n] = T[n] + n = T[n+1] - 1, from n = 1, where T[n] is the nth triangular number. The sum of the nth row is 3 * 2^(n-1). Numbers down the central spine are multiples of 3, and they grow at the same proportional rate as those of Pascal's triangle (ignoring line 0): a[1] = 3; a[2] = 3 * 6/2 = 9; a[3] = 9 * 10/3 = 30; a[4] = 30 * 14/4 = 105; etc. a[n] = (3/2)(2n)!/(n!)². While this triangle cannot be used to find binomial expansion coefficients, its ascending shallow diagonals can be used to find polynomials for L[2n], L[3n], L[4n], etc. For example: L[3n] = 1 * L[n]^3 - 3 * (-1)^n * L[n]; ''''' ''''' L[4n] = 1 * L[n]^4 - 4 * (-1)^n * L[n]^2 + 2. ''''' ''''' ''''' In short, this variation on Pascal's triangle is packed with interesting properties, including some that it shares with its better-known cousin. For example, the hockey stick pattern applies from either edge, as it does in any such triangle: e.g., 2 + 5 + 9 = 16; 1 + 4 + 9 +16 = 30. Other variations yield different pairs of Fibonacci-type (Lucas-type) sequences and other more obscure sequences. For example: 1 3 (Σ = 4 = 4 * 1 = 2^2) 1 4 3 (Σ = 8 = 4 * 2 = 2^3) 1 5 7 3 (Σ = 16 = 4 * 4 = 2^4) 1 6 12 10 3 (Σ = 32 = 4 * 8 = 2^5) 1 7 18 22 13 3 (Σ = 64 = 4 * 16 = 2^6) 1 8 25 40 35 16 3 (Σ = 128 = 4 * 32 = 2^7) 1 9 33 65 75 51 19 3 (Σ = 256 = 4 * 64 = 2^8) This variant contains: the natural numbers from 3; the sequence 1, 4, 7, 10, 13, ..., 3n + 1, from n = 0; the sequence 3, 7, 12, 18, 25, 33,..., n(n + 5)/2, from n = 1 (again similar to triangular numbers); the pentagonal numbers, 1, 5, 12, 22,..., n(3n - 1)/2, from n = 1; the pentagonal pyramidal numbers, 1, 6, 18, 40, 75,..., n²(n + 1)/2, from n = 1; the Lucas numbers again (from L[2], by summing the shallow descending diagonals); the sequence, 1, 4, 5, 9, 14, 23,... (another Fibonacci-type sequence, sometimes called the Pibonacci numbers, by summing the shallow ascending diagonals); and so on. The sum of the nth row is 4 * 2^(n-1), or 2^(n+1). Numbers down the central spine are multiples of 4, and once again they grow at the same proportional rate as those of Pascal's triangle: b[1] = 4; b[2] = 4 * 6/2 = 12; b[3] = 12 * 10/3 = 40; b[4] = 40 * 14/4 = 140; etc. b[n] = 2(2n)!/(n!)². I do hope somebody appreciates the beauty of this. It's amazing what one can find just by looking for patterns. Please feel free to point out any errors.
Hi Matt, great video! Also, I am looking at your sideburns and I sense that they want to take it to the next level. Come on Matt, history is ripe for the return of epic sideburns; lead the way!
Great video, we never made that link between binomial expansion and coefficients in school and were instead unfortunately just told to learn it. The audio on this video was unfortunately not great, sounding far too bass heavy on a speaker setup but that's the only thing that needs improving.
Couldn't you say that (briefly because I'm typing on a device and I'm lazy) the connection is the choose function produces the numbers of combinations of items chosen from sets. And a binomial expansion requires all the possible combination of the terms with their different exponents which is all the possible combinations of item's in a specific set.
Is there a name for the series of mathematical functions that begin with addition, multiplication and powers? There's a pattern in how these follow each other, taking 7 as an example. 7+7=7*2, and 7*7=7^2. So is there a notation that would go between 7 and 2 for 7^7? And if there is, for example #, and 7^7=7#2, is there a name for the next step of 7#7 which can be written 7something2? Does this series of functions have a name? Is it useful for anything in mathematics? I only have a basic knowledge of mathematics but I haven't found a name or anything about the properties of this series. I can only imagine that numbers using any of these higher functions would get extremely large extremely fast.
Matt, I'm firmly convinced that if you had been my math teacher in school, I would have believed I actually liked math. Unfortunately, I believed I loathed math for most of my life, even though I almost always got good grades in it.
I don't know if it is a matter of cause and effect but it seems that whenever you are following a path along into the inner part of the triangle, leaving the 1's, the end of the path is the number of possible paths leading to the number.This holds true only if your path consists only of downward-lefts dl and downward-rights dr. For example 6: dr-dr-dl-dl; dl-dl-dr-dr; dr-dl-dl-dr; dl-dr-dr-dl; dr-dl-dr-dl; dl-dr-dl-dr.
+Alexander Nauditt That's because all these problems (choosing balls, binomial coefficients, paths down a graph and many others) are all different applications of the same principle of counting the different ways to pick m elements out of a set of n. In your triangle path example: - each step goes down one line, so all paths going to the 6 are 4 steps long, - each step is either a dl or a dr, - a dl-dr and a dr-dl path are equivalent so you don't care about the order in which the dl's and dr's are picked in your path, you just need that the path contains 2 dr's, So the number of paths going to the 6 is the number of paths of length n=4 that contain exactly m=2 dr's. That's 4 choose 2. Almost any counting problem can be reworded as picking m elements out of n or as a combination of several of these (n1 choose m1, then n2 choose m2, etc). And that's why binomial coefficients appear so often in maths and physics.
7:01 - You forgot to add your Σ expression to the start of that. Here's the full expression for your expansion of (a+b)^n: n Σ (n nCr m) × a^(n-m) × b^(m) m=0
i hope matt could do videos that could help alevel students such as differentiation and integration by first principles. binomial is alevel so he could do this entertainment for everybody but it can also help students across the world
There are many more interesting things about the binomial or choose function. For example n choose k also represents how many subsets with k elements there are in a set with n elements. Then there's the interesting Star of David theorem. And also the Catalan numbers.
Great channel! I have a question: Why do Gaussian Bells (or Normal Distribution Bells) work so well? Or, more casually: Why do mathematicians use that particular kind of curve for continuous probability? Why not a semicircle?
In my proof pascals triangle is zero indexed. This means that the top of the triangle is in row 0, and the leftmost position is 0. In row n there are n + 1 positions, where the first position is 0 and the last position is n. To prove that pascals triangle is composed of the binomial coefficients, I will introduce the mathematical formula for the choose function, nCr, in terms of the number of object you choose from, n, and the objects you choose, r. The function is n!/(r!*(n-r)!). The reason the function looks like this is because you can imagine choosing r objects from a pool of n as this procedure: Sort the n objects, which can be done in n! ways. You then draw a line to separate the object you choose from the ones you don't choose. You are not interested in the order of any of the two subsets, so you divde by the amount of ways to sort each of these subsets, which are r! and (n-r)! respectively. The 1s in the triangle can be shown to fit because nC0 and nCn are both equal to 1 as per the formula. Now, if pascals triangle is composed of binomial coefficients, then the coefficient in row n and position n, n!/(r!*(n-r)! should be equal to the sum of the coefficients above it, which are both at row n-1, one is at position r and one is at position r-1. So if the statement is true then n!/(r!*(n-r)!) = (n-1)!/(r!*(n-1-r)!) + (n-1)!/((r-1)!*(n-r)!). We need some restrictions on n and r. We are only concerned about the numbers that are not 1s anyway, so n > 1 and 0 < r < n. These restrictions make certain multiplications legal, because they would otherwise multiply by 0. [insert maths operations] and since the sum can be show to be equal to the left hand side in the above equation, we can say that the statement is fulfilled.
is it also linked to, let's say, last coefficients of squares from 1 to 9 being symmetrical like Pascal's triangle? By the way great lesson and fun as always.
Hey Matt, I was reading up on Nuclear Magic Numbers. The wiki says that they are derived from binomial co-efficients but I don't see how their formula produces the numbers. Maybe there is an interesting video in that. 2, 8, 20, 28, 50, 82, and 126
Is there a reason why the (get ready for some terrible terminology) ending parts of the equation's exponents always equal the original power it was brought to? For example, in your (a+b)^4 example, you get a^4, a^3b, a^2b^2, ab^3, and b^4. Each exponent, when added together, makes 4 (4=4; 3+1=4; 2+2=4; 1+3=4; 4=4). It doesn't quite make sense to me... I type this all out, then I realize why "sigh". It's because of the rule of when to add exponents, namely during multiplication. By virtue of the fact that we are multiplying 4 times per step (I guess is the right word?), we will get 4 items multiplied. Little derp.
+Graham Rich because you have four numbers multiplied together, each number has exponent 1 and if you add the exponents of four numbers you of course get 4. if you have (a+b)^n you would get something like (a+b)(a+b)(a+b)......(a+b) and in totalt there would be n parentheses. When you choose one number from each you effectively choose n numbers, all with exponent 1, and when you add 1 n times the result is n.
+standupmaths I don't know if you're gonna get this, but if you do. How do I calculate these? Looking at a task where they ask how many ways you can write 6 digits in ascending order from left to right. To do so you just write the numbers 1-9 and cross out 3 of them which can be done in 9 choose 3 ways, and then I have to calculate (a+b+c)^9 , right?!? And just how on EARTH am I going to do that?!?!
I noticed the same thing in the main group of elevators at my office -- elevator D has "Elevator No. D" printed on the inside, next to the door (and likewise for the other elevators).
Hey Matt I have a question you might be able to answer in your next Matt Explains (BTW loving this series!) So we doing logarithms in class and there is this basic rule saying that a^x=b LOG(base a) b = x. So I wondered that LOG(base -2) -8 = 3 So I put it into my calculator and to my surprise I got an answer with an imaginary number. I asked my teacher and she said that it was because logs are based on a table and negative numbers are not on a table. That answer did not and still does not satisfy me. So could you please explain it. Many thanks
Maybe, what she meant is, that your calculator works with tables, which have the logs in them, because it is an easier way to figure them out. (I don't know whether this is true). Maybe she just didn't want to say, that she didn't know the answer right ahead. People get embarrassed of these kinds of things. I don't know, how far you have come in your maths class, maybe the whole try of an explanation following in my next comment is way to far reaching. I try my best to keep it easy, please tell me, if you did not understand it and what of it.
As far as I know (math student, 3rd semester, so you could basically say I'm making wild guesses with the help of wikipedia), the logarithm can't be defined with negative real numbers as *base.* (Though it is possible to find things like log(base 2) of -4 ) You will always find an exception and things you can't display with real numbers. What's the logarithm of -4 base -2? Btw, what was the imaginary number, your calculator gave you? Have you tried multiplying it? Just do a few experiments yourself. I will try to give you a few examples and hints, please notice that I cannot answer your question easily... The logarithm is, as you mentioned, defined as the reversal of raising to a power. Maybe you have already learnt the rules for logarithms and the natural logarithm? Then you know, that log(base a)b = log(base c)b / log(base c)a e.g. log(base 2)16 = log(base 4)16 / log(base 4)2 or 4 = 2 / (1/2) Ah yes... you can do partial exponents, too. It's similar to the root of something. _(Square root(2) = 2 ^ (1/2) )_ In fact you can have all real numbers as exponents and define a power function for it. (Do you already have functions? It's when you assign a value to a variable, following a certain rule. E.g. y = 2 * x ; I hope you do...) So you can figure out things like log(base 2)3 and get real numbers. Let's use y= 4^x and y=(-4)^x as an example. log(base 4)16 = 2 log(base -4)16 would logically be 2, too. What about this? log(base 4)2 = 1/2 , because square root (4) = 4^(1/2) = 2 log(base -4)2 = ? What is the square root of -4? Well sqr(-4) = sqr(-1 * 4) = 2 * sqr(-1) = *2 * i* _You get an imaginary number._ As soon as you start to insert negative values into your exponential funktion, you will get imaginary numbers for every partial exponent. Take a look at the graph at the top of this website: www.wolframalpha.com/input/?i=%28-2%29^x (Wolfram Alpha or similar sites are very useful, once you get into the more difficult parts of maths, they visualise things quickly) I hope I could help you at least a little and did not just confuse you with the many concepts you have not yet been taught. Stay curious! Lia
Thank you very much for the answer but I am still not extremely clear on how log(base -2) -8= 1.09 -0.42i (this is a rounded value from my TI calculator)
I got a question that I would like you to explain. First since a and b in the binomial are arbitrary that would mean that the coefficients are arbitrary, meaning there are the same number of ways to choose 3 a's and 1 b than 1 a and 3 b's. Why does this symmetry occur.
+qwertyuiopzxcfgh. Yeah, I first saw it on my calculator as nCr and was using it for puzzles, (like counting presents in the 12 Days Of Christmas song, or generalising 'triangular numbers' up to 'tetrahedral numbers' etc), long before it was introduced in school, and I just thought it stupid how they wrote it as a 2 vector.
Yes, I'm familiar with the C notation. I actually like the ( ) notation because it gives the function an identity not directly linked to choosing (just one of its many uses).
Can you systematically derive the binomial theorem from something that is so true and obvious as to be trivial? I understand how the formula works, but I don't understand how we got there in the first place. Also, I don't understand how we can be sure that the formula holds for all possible values of n.
I'm working up to the challenge of a good "why" video. Too many mathematics videos are content with pointing something amazing out but then never explaining it.
![Matt Explains: The Lottery [featuring: Choose Function, Infinite Geometric Series]](http://i.ytimg.com/vi/lP58mP8Wchc/mqdefault.jpg)
Best guess at the Pascal's Triangle:
This goes back to what you were doing at the beginning, expressing each binomial expansion as the previous one multiplied by another (A+B). In that system, each of A and B gets multiplied by each of the previous parts, and the sum of all those results gives us our next expansion.
So let's look at the third expansion: A^3+3(A^2)B+3A(B^2)+B^3. To find our fourth, each of those gets separately multiplied by an A and a B. To simplify, let's look at that with just the first two terms: (A^3+3(A^2)B)*(A+B).
If we multiply A^3 by A, we get A^3. If we multiply it by B, we get (A^3)B. If we multiply 3(A^2)B by A, we get 3(A^3)B. If we multiply it by B, we get 3(A^2)(B^2). Now the question is, how many (A^3)B results did we get, total? Well, we got one for every A^3 in the previous expansion (1 total), plus one for every (A^2)B in the previous expansion (3 total). No other variable segment can produce (A^3)B, so our result is the sum of those two terms. Since each variable segment can only become one of two different variable segments in the next expansion, this should happen with any other pair of terms as well. (This one also gave us a bonus term because the A^4 term can only be approached by one segment, but that won't happen with more central segments.)
That took a lot of work to figure out. I should do this stuff more often...
Now I just need a way to explain the concisely!
standupmaths Isn't it because of nCr + nC(r-1) = (n+1)Cr ?
As soon as I saw the 4641 I though Pascal’s Triangle must be coming. I use the first 4 rows when I want an example for some compound interest in class and just use 10%.
I swear to god that if I'd had you as an AS Maths teacher instead of the disinterested person I did get, I wouldn't have miserably failed it haha. Another superb video, entertaining and well laid out :) Thanks, and I hope to keep seeing em coming!
It's never too late to learn AS maths! Or at the very least, rediscovering a love of mathematics.
So nice of you to say that my comment was your favourite =)
Regarding Pascal triangle it is pretty easy to see why it works
In binomial coefficient we have for example
1 3 3 1
Then when we multiply by (a + b) if we consider the result it will be
1 3 3 1 0
+
0 1 3 3 1
=
1 4 6 4 1
Because powers will be shifted by a or b, so we sum up previous coefficient with itself but shifted by 1 position. And pascal triangle does exactly that, you sum up (k,n) position with (k,n+1) position in order to get the position (k+1, n+1) (where k is the number of row in Pascal triangle), and positions of (k, 0) and (k,k) are always 1 (which even works for 0's row, because 1 = 1)
P.S. Binomial Coefficient is one of my favourite subjects though, I was so interested in this concept that I actually asked my dad (who has PhD in Physics) to explain it to me when I was in 3rd Grade... Though I asked him to explain the Choose Function back then (even though I didn't know the word "function" back then), only later on I found out about Binomial Coefficients and Pascal Triangle. And I personally prefer "Binomial Coefficient" or "Newton's Binomial Series" names.
It was a great comment! Thanks for your contributions to my channel. I'll try to get some of this in my next video.
I noticed the Pascal's triangle thing when you wrote the numbers down, but I couldn't figure out why. I'm really looking forward to you explaining why :)
Hopefully the wait will not be too long!
+standupmaths there's only one now, your sideburns must have scared them off.
+Harm Prins Hint: You can already find the answer in this video if you look at the way terms are organized in the computation of (a+b)^4 from (a+b)^3 x (a+b).
Harm Prins the explanation that seems most likely to me is that each number on Pascal's triangle lists the number of ways to reach that location from the top of the triangle, moving downward through adjacent numbers.
there's only one way to reach the top number: by going nowhere.
there is only one way to reach the first number on the second row, which is by moving one down to the left
there is only one way to reach the second number on the second row, which is by moving one down to the right.
there are two ways to reach the second number on the third row. one is by moving right, then left. the other is by moving left, then right.
and so forth.
I'm sure there's a more rigorous proof behind that, but that seems to be the gist of it
standupmaths Still waiting...
Pascal's triangle answer: the number of paths you can take from the top to a number. i.e. you can take 6 different paths to get from the 1 at the top to that number 6.
What a fun way to wake up on this post finals day. :D Thank you very much, Matt!!
Then it's a different path altogether.
Please do MORE!!
Can DO.
@@standupmaths Reporting in: Has done!
It's clear from the definition of the choose function that for any natural number n,
C(n, 0) = C(n, n) = 1.
It is relatively straightforward to show - using the definition of the choose function and some algebra - that for any natural number m such that 0
Matt you're a legend. I've been struggling with this for the past month and I finally understand.
Thank you so much for explaining this concepts and connection. I wish every teacher in the U.K teaches maths like you did in this video .
Or every teacher in America for that matter.
At this point i have to:
Zeroth: wander what the hell will happen if you make a triangle with trinomial coefficients.
1/2th: sugest it has to do with "Pascal's tetrahedron"
First: admit you've blown my mind
Second: thank you this awesome video
Third: point out that at the time I'm writing this video has 234 likes and 2 dislikes
Fourth: you're welcome. Thanks for watching!
Since the moment I ended writing my comment I started thinking about the zeroth point, and I foun that not only I was right in the 1/2th point but this is true for any given dimension. I've tryed it for the first five dimensions and I'm amazed of how quickly they grow in size. I was also very glad about knowing that all the time I spent calculating and writing Pascal's n-simplex down in my window.
Love these videos so far, it's like having extra Numberphile videos to watch! I just wanted to ask if I may, do you have an intuitive understanding of maths in any way or is it something you've had to work extra hard on? I ask because while I love maths I really struggle with it, which is why I love videos like this. Makes it all that bit easier to grasp.
Glad you enjoy them! Experience can cause you to grow something a bit like mathematical intuition, but I don't think it ever becomes second-nature.
Another truly *standup video!* Splendid! You never sat down once!
I was breathlessly waiting for you to show the direct connection between Choose and Binomial by showing the Choose function being used to 'assemble' each term in the expansion.
And son-of-a-gun if you didn't do exactly that! Kudos, Matt, and thanks!
Fred
I had a few things to do today... but instead I spent most of the day watching your videos. Those are truly awesome :)
When you are at 121 it means that you have 1 way to have aa 2 ways to have ab and 1 way to have bb. When you start the 1331 the first 1 is the number of ways to get aaa, which is the 1 way to get aa with an extra a. The first 3 is the number ways to get aab, and that is the 1 way to get aa with an extra b plus the 2 ways to get ab with an extra a. And that is - by example - why you can get the coefficients by adding pairs of the coefficients for the previous power.
Most of video: *"Huh, pretty cool that it works like that. Makes sense."*
9:45: *MIND->BLOWN.*
Awesome video matt! - thank you for sticking with it,
and don't forget to do a standupmaths about pascal's triangle.
I'm thinking of doing a whole Pascal's Triangle series. We'll see.
Apart from the fact that I have this geniunely funny guy explaining Math to me, I'm also satisfied that he cares to explain "why". This is something noone does.
I don't care much about the maths, but I watched the whole damn video because I love your passion and enthusiasm. Would gladly watch more. Wish I had someone like you as a school teacher all those years ago.
Yes, there is no match for an enthusiastic teacher. But it's never too late to learn maths!
For every number in the triangle, N is the number of ways to get back to the top, moving either left or right for every tier on the way up. For the 1s on the edges, they have no choice but to keep going on the edge. The first 2 can go either right-left or left-right. The first 3 can go left-right-right, right-left-right, or right-right-left. The first 4 can go left-right-right-right, right-left-right-right, right-right-left-right, or right-right-right-left. That's all boring because it's on the edges while there is some "choice" in the order of the rights and lefts, you still only choose one left. Hence, for the 4 it is 4 choose 1 or 4C1. For the 6, now, it can go R-R-L-L, R-L-R-L, R-L-L-R, or the mirror images of all of those, since it's in the middle. Here, we are picking 2 lefts, so it is 4C2. Then it's 4C1 again and finally 4C0.
As to _why_ these numbers are the way to get back to the top, I'd wager it's because every extra layer is basically another "term" of the binomial where you ought to choose a or b, in this case right and left. As you add a layer, terms build up, and so the layer inherits the terms of the previous one. And adding the two numbers above is adding the two paths to get back to the top from each of those instances. The 6 has two ways up - the two 3s. Each of those 3s has 3 paths up, so 3 + 3 = 6.
So there you go. The Pascal Triangle works because you add the two numbers of paths above it, since you need to choose one out of two of them. And that's nested, and as we know, nested things tend to make fractals, and oh dear it seems Pascal's triangle has a link to the Sierpinski triangle ;)
And Matt, if you're reading this, hi!
This is indeed pretty much it! I just need to find a concise way to explain it…
+standupmaths Maybe using recursion? T(n,x) = T(n-1,x-1) + T(n-1,x) with T(0,0) = T(0,n) = 1
+standupmaths Instead of explaining why it relates to the choose function directly, show that adding a row to the triangle is a skewed version of multiplying by 11. 11^n = (10 + 1)^n, so you have binary coefficients multiplied by powers of 10. You've already proven binary coefficients are related to the choose function, so the numbers in Pascal's triangle are too. QED.
Imagine if this guy was my Maths Teacher. So good!!!! :) :)
I was once a real maths teacher…
+standupmaths
Seriously. Come teach me, because my maths teacher is the worst. I love maths, but she ruins the lessons I have. Also, I have a test tomorrow. :(
@@tommysadler1033 f
Wait till you meet salman khan of khan academy
Yaaaaay! Matt is the best. I absolutely positively love Pascal's Triangle, & it overjoyed me in my sophomore (10th grade) algebra class when I learned that you could find binomial coefficients with it. Thank you, Mr Parker! :)
So Long & Thanks For All the Fish,
Lawrence Calablaster
Great video Matt, forget about the video strobing, it's fine.
It's the content that matters, and yours is great.
I'm a college student (year 13 UK) and I just want to say I love your videos and that I just did this in class Monday
I found this by chance and it has really helped thanks for the great videos
The pascal's triangle works because you are multiplying the previous term including the corresponding variables by x and y. For example, multiplying x³y² by x gives you x⁴y² but multiplying it by y gives you x³y³. But you could also get x⁴y² by multiplying x⁴y by y, and you could also get x³y³ by multiplying x²y³ by x. This is why each number gets added twice, once to the number on the left, and once to the number on the right. This also explains why if you add up all the numbers in each row, it will give you increasing powers of two as you go down. Because each number is getting added twice, its the same as multiplying everything by two.
Yes, I think the rows summing to powers of two is a nice surprise but obvious once you think about it.
+standupmaths Huh, I'd never noticed that, but yeah that makes perfect sense.
Brilliant explanation Derek C.
Video tips: the best way to fix the focus problem is (1) forget autofocus and use manual focus on the whiteboard and (2) increase the lighting so that your camera uses a smaller aperture. If you want better audio you could also wear something with a higher neckline and get the mic closer to your mouth, without obstruction from clothing... Great video though! I kept wondering why you hadn't mentioned Pascal's triangle... and then you did :)
Yes, I do have a bit more space for the camera but I'll need to get better lighting. And yes, a better-placed lapel mic.
Each position is the number of paths to that position from the beginning if every movement is a downward diagonal, like checkers. Which is essentially choosing a number of left and right moves, a and b moves, which always have to add up to the same amount to get to the same position, all the ways to get 3 Lefts and 4 Rights, which is the choose function.
Which isn't the Why the numbers are that so much as the surface connection I'll probably come back and have a go at that later.
That is an amazing start! I plan to use the path example in my video.
dude you are awesome. I love the videos you put on UA-cam. Really I myself want to be a mathematician. Great video. Loved it.
Explanation at 9:08 was initially very confusing for me because the situation seemed more like it required permutations (P) rather than choose (C). Perhaps it is useful to imagine the a or b values as being separate for the purposes of the explanation, which can be done through assigning a subscript value. Then the reason why the n value in n choose r equals the power of the binomial is because this power of the binomial is the same as the number of "separate" a or b values which can be picked. Other than that, thanks for the great video!
Really helps fill in the small gaps of C2 A level. Loving the series
C2 is great fun. But A2 really takes off!
I'm from indian you are great explanation of this topic
Keep it up ALWAYS
You know ours mathematicians. Like RAMANUJAN. ARYABHATTA. SAKUNTALA DEVI
All are genius
You also like that
Thanks to give your time to read my comment
I salute you
Aw yiss. I used to always expand binomials with Pascal's Triangle. It used to saved me lots of time in exams. Ah, fond memories.
No, Matt, we're not your people... You man are OUR people!
I cant believe I watched this video when it only had 18 views! I'd like to say that's a good explanation, and thank you for spending your time making awesome videos!
You are ahead of the wave! Thanks for being awesome and watching my videos.
Here it is! :
Each number in Pascal's triangle is the ammount of paths you can take to get up. So why can you calculate a number by adding the two above it? because after you choose one of them (let's say the number's 84), there are 84 paths to continue after that, so if you add it up with the paths to the other option you get all the paths for that number.
Starting at the top (it's easier to visualize), to get to any position in the triangle you need a very specific ammount of turns to the left and to the right, because anything else would leave you in another square; however, these turns can be in any order and you'll still get to the right position.
So, let's start by analyzing the leftmost side: for all steps you need 0 right turns and S (the number of the step counting from the top) left turns. S choose 0 returns 1 for any S, so that's why all numbers in the left are 1s
After you have that it's really easy to continue: the second squares from the left require one step to the right, so you do S choose 1 wich varies depending on S; the next square requires 2, the next 3 and so on... You can also count the number of left turns it needs and get the same result because of symmetry, or count either the right or left turns starting from the right.
So from what i've explained above, we get this:
If you're in the Nth step (from top to bottom), you calculate the Mth number (from left to right, anything from 0 to N) by doing N choose M, or choose(N,M) as i prefer to say it (because programming)
Oh! also,adding up the numbers in a step gets you a power of 2 because you can express all L/R paths as binary numbers! and each step uses all possible paths.
As you go down the triangle, you can go either left or right. So, for the 1,4,6,4,1 line, of the 4 steps you go down, you would need to go left twice and right twice to get to 6, i.e 4C2
I love it. Thanks for continuing to do these. Looking forward to more!
Also, if you cut parallel lines at a jaunty angle through Pascal's triangle, you get the Fibonacci sequence.
1=1
1=1
1+1=2
1+2=3
1+3+1=5
1+4+3=8
And so on
Pretty sure i got it! i'll write it later because i can't now but it relates to binary and how many ways there are to arrange a certain number of ones or zeros (using the choose function)
Just woke from a dream where i was trying to explain to someone the connection between pascal's triangle and factorising polynomial functions to solve trivial problems in Asymmetric cryptography and now ive woken up and here i am. Thanks for the reminder :)
Incredible sideburns. Terrific vid as well. But dang, those sideburns.
I know there tons of numberphile videos about this, is 1 + 2 + 3 + 4 + 5... = -1/12? Or is there something wrong with this proof. I would like to see more videos about infinite sums. I love how you keep making more videos. I thought this channel was going to go dead but now there are more Math videos. YAY!
The -1/12 thing is a world of confusion but I might take it on one day. And yes: I'm trying to keep this channel alive! Hard to find the time+money though.
I have a TI-30XS MultiView calculator that has a "table" button, with which I can input an expression with one variable (x), and it returns a list of appropriate y values.
I can't use the proper notation, but instead I have to use things like "n nCr x. This works well for the rows of Pascal's Triangle, but I found a way to generate the sequences of numbers in the diagonals of the Triangle: if I use x nCr n (for example, x nCr 2) I get the triangular numbers. If I use x nCr 3, I get the tetrahedral numbers, etc.
This works for any of the diagonals of the Triangle! 🤓
Can you please please please make a video about trig identities? I wanna know why sin(a+b) is equal to sinx cosy + siny cos x and same to cos and tan addition. You and Numberphile are the only channels that makes math sound exciting. Thank you
Try Kahn Academy. They have some great videos on this.
Thanks 😎👍
I now know why there is a link between the combination function & binomial coefficient
Hooray for math! Thanks Matt
The pascal's triangle at the end just made everything click firmly into place.
Videos like this one are the main reason I don't hate math. Schools should teach fun things like this.
Sadly school teachers have to cover the curriculum which leaves very little time for anything else like this. And if a teacher does make time, students/parents complain that it's not going to be on the test.
The pascal's triangle gives a count of the number of shortest paths from the top to the current number.
Why does this happen: Each number has two "parent" numbers (or one in case of the outside numbers of the triangle). The only possible way to arrive at the current number (on a shortest path) is via one of these parents. If the parents number display the number of paths to their number, adding up these parents will result in the amount of paths to the current number. This reasoning can be from the bottom up to the second row of the triangle (at 1 1) . Because these 1's are the number of paths to these numbers (both 1's can only be reach in one way), the theory is correct. For all the numbers with only 1 parent, there is only 1 shortest path to reach these numbers and they therefore always contain a 1.
In short each number has two parents with a count of the total number of paths to these parents, by summing them the total number of paths to the current number is obtained.
Why does this give the correct result: If we take a look at the 5th row of the triangle one path two the 6 could be described by LRLR, in which the L and R give a choice to either the left or the right respectively. The path would in this way follow the numbers: 11236. To arrive at the 6, precisely two of the 4 choices must be Right (the other two automatically being a Left) or in other words you have 4 options and you have to choose 2 correct ones out of it, which brings you right back to the start of the video (the choose function/binomial coefficients).
Interesting observation, nicely explained
Hi, Matt. Perhaps this will interest you.
A variation on Pascal's triangle yields both the Fibonacci sequence from F[3] (2, 3, 5, 8, 13,...) and the Lucas numbers from L[1] (1, 3, 4, 7, 11,...), by summing the descending and ascending shallow diagonals respectively:
1 2 (Σ = 3 = 3 * 1)
1 3 2 (Σ = 6 = 3 * 2)
1 4 5 2 (Σ = 12 = 3 * 4)
1 5 9 7 2 (Σ = 24 = 3 * 8)
1 6 14 16 9 2 (Σ = 48 = 3 * 16)
1 7 20 30 25 11 2 (Σ = 96 = 3 * 32)
1 8 27 50 55 36 13 2 (Σ = 192 = 3 * 64)
This triangle also yields: the natural numbers from 2; the odd numbers; the squares; the square pyramidal numbers; numbers of the form n(n + 3)/2 (similar to triangular numbers); the sequence S[n] = T[n] + n = T[n+1] - 1, from n = 1, where T[n] is the nth triangular number.
The sum of the nth row is 3 * 2^(n-1). Numbers down the central spine are multiples of 3, and they grow at the same proportional rate as those of Pascal's triangle (ignoring line 0):
a[1] = 3; a[2] = 3 * 6/2 = 9; a[3] = 9 * 10/3 = 30; a[4] = 30 * 14/4 = 105; etc.
a[n] = (3/2)(2n)!/(n!)².
While this triangle cannot be used to find binomial expansion coefficients, its ascending shallow diagonals can be used to find polynomials for L[2n], L[3n], L[4n], etc. For example:
L[3n] = 1 * L[n]^3 - 3 * (-1)^n * L[n];
''''' '''''
L[4n] = 1 * L[n]^4 - 4 * (-1)^n * L[n]^2 + 2.
''''' ''''' '''''
In short, this variation on Pascal's triangle is packed with interesting properties, including some that it shares with its better-known cousin. For example, the hockey stick pattern applies from either edge, as it does in any such triangle: e.g., 2 + 5 + 9 = 16; 1 + 4 + 9 +16 = 30.
Other variations yield different pairs of Fibonacci-type (Lucas-type) sequences and other more obscure sequences. For example:
1 3 (Σ = 4 = 4 * 1 = 2^2)
1 4 3 (Σ = 8 = 4 * 2 = 2^3)
1 5 7 3 (Σ = 16 = 4 * 4 = 2^4)
1 6 12 10 3 (Σ = 32 = 4 * 8 = 2^5)
1 7 18 22 13 3 (Σ = 64 = 4 * 16 = 2^6)
1 8 25 40 35 16 3 (Σ = 128 = 4 * 32 = 2^7)
1 9 33 65 75 51 19 3 (Σ = 256 = 4 * 64 = 2^8)
This variant contains: the natural numbers from 3; the sequence 1, 4, 7, 10, 13, ..., 3n + 1, from n = 0; the sequence 3, 7, 12, 18, 25, 33,..., n(n + 5)/2, from n = 1 (again similar to triangular numbers); the pentagonal numbers, 1, 5, 12, 22,..., n(3n - 1)/2, from n = 1; the pentagonal pyramidal numbers, 1, 6, 18, 40, 75,..., n²(n + 1)/2, from n = 1; the Lucas numbers again (from L[2], by summing the shallow descending diagonals); the sequence, 1, 4, 5, 9, 14, 23,... (another Fibonacci-type sequence, sometimes called the Pibonacci numbers, by summing the shallow ascending diagonals); and so on.
The sum of the nth row is 4 * 2^(n-1), or 2^(n+1). Numbers down the central spine are multiples of 4, and once again they grow at the same proportional rate as those of Pascal's triangle:
b[1] = 4; b[2] = 4 * 6/2 = 12; b[3] = 12 * 10/3 = 40; b[4] = 40 * 14/4 = 140; etc.
b[n] = 2(2n)!/(n!)².
I do hope somebody appreciates the beauty of this. It's amazing what one can find just by looking for patterns. Please feel free to point out any errors.
choosing n-1 is the same as choosing not 1
+Dan Dart And that explains why the binomial coefficients and Pascal's triangle are symmetrical.
indeed, choosing n-a is the same as choosing a.
I was going to like this video, but there were 111 likes and 1 dislike and I didn't want to be a combo breaker. Great video, anyway! :)
Those days are now long gone. As I type: two dislikes!
+standupmaths miraculously, one of those dislikes has disappeared!
+standupmaths It's 1 dislike again. Weird. Huh??
Hi Matt, great video! Also, I am looking at your sideburns and I sense that they want to take it to the next level. Come on Matt, history is ripe for the return of epic sideburns; lead the way!
I am teetering on the sideburn edge, afraid of what is on the other side (burn).
I think it would be great to see Matt do a combinatorics word problem. We did quite a few of them in my Descrete Math courses.
+standupmaths Isn't the choose function same as combinations or nCr(side note: n is superscript & r is subscript)which is = n!/[r!(n-r)!]
Matt, you are awesome, keep it up and great vid :)
Great video, we never made that link between binomial expansion and coefficients in school and were instead unfortunately just told to learn it. The audio on this video was unfortunately not great, sounding far too bass heavy on a speaker setup but that's the only thing that needs improving.
Yes, I think the audio quality might be the next thing I try to fix.
Couldn't you say that (briefly because I'm typing on a device and I'm lazy) the connection is the choose function produces the numbers of combinations of items chosen from sets. And a binomial expansion requires all the possible combination of the terms with their different exponents which is all the possible combinations of item's in a specific set.
Again very entertaining. Keep up the good work!
I'm supposed to be doing homework...but I like this...video's over. I'll go do homework now.
Is there a name for the series of mathematical functions that begin with addition, multiplication and powers? There's a pattern in how these follow each other, taking 7 as an example. 7+7=7*2, and 7*7=7^2. So is there a notation that would go between 7 and 2 for 7^7? And if there is, for example #, and 7^7=7#2, is there a name for the next step of 7#7 which can be written 7something2? Does this series of functions have a name? Is it useful for anything in mathematics? I only have a basic knowledge of mathematics but I haven't found a name or anything about the properties of this series. I can only imagine that numbers using any of these higher functions would get extremely large extremely fast.
+Conway79 Knuth's up-arrow notation
Matt, I'm firmly convinced that if you had been my math teacher in school, I would have believed I actually liked math. Unfortunately, I believed I loathed math for most of my life, even though I almost always got good grades in it.
I don't know if it is a matter of cause and effect but it seems that whenever you are following a path along into the inner part of the triangle, leaving the 1's, the end of the path is the number of possible paths leading to the number.This holds true only if your path consists only of downward-lefts dl and downward-rights dr.
For example 6: dr-dr-dl-dl; dl-dl-dr-dr; dr-dl-dl-dr; dl-dr-dr-dl; dr-dl-dr-dl; dl-dr-dl-dr.
+Alexander Nauditt That's because all these problems (choosing balls, binomial coefficients, paths down a graph and many others) are all different applications of the same principle of counting the different ways to pick m elements out of a set of n.
In your triangle path example:
- each step goes down one line, so all paths going to the 6 are 4 steps long,
- each step is either a dl or a dr,
- a dl-dr and a dr-dl path are equivalent so you don't care about the order in which the dl's and dr's are picked in your path, you just need that the path contains 2 dr's,
So the number of paths going to the 6 is the number of paths of length n=4 that contain exactly m=2 dr's. That's 4 choose 2.
Almost any counting problem can be reworded as picking m elements out of n or as a combination of several of these (n1 choose m1, then n2 choose m2, etc). And that's why binomial coefficients appear so often in maths and physics.
Correct me if I am wrong. at 7:00 this formula have error. it should be a power n and b power n-m. as per the link given in description.
the moment around 8:30 where i feel warm in my heart and start to smile
7:01 - You forgot to add your Σ expression to the start of that. Here's the full expression for your expansion of (a+b)^n:
n
Σ (n nCr m) × a^(n-m) × b^(m)
m=0
i hope matt could do videos that could help alevel students such as differentiation and integration by first principles. binomial is alevel so he could do this entertainment for everybody but it can also help students across the world
This channel is more awesome by the video!
awesome video! really helped me understand!
Hurray! That's my goal.
There are many more interesting things about the binomial or choose function. For example n choose k also represents how many subsets with k elements there are in a set with n elements.
Then there's the interesting Star of David theorem.
And also the Catalan numbers.
I am a bit scared of starting a series on interesting things in Pascal's Triangle because it may never end.
man thank you for these. I've always liked math, not good at it tho. But you really make me want to do some math haha
Great channel! I have a question: Why do Gaussian Bells (or Normal Distribution Bells) work so well? Or, more casually: Why do mathematicians use that particular kind of curve for continuous probability? Why not a semicircle?
great vid Matt!
In my proof pascals triangle is zero indexed. This means that the top of the triangle is in row 0, and the leftmost position is 0. In row n there are n + 1 positions, where the first position is 0 and the last position is n.
To prove that pascals triangle is composed of the binomial coefficients, I will introduce the mathematical formula for the choose function, nCr, in terms of the number of object you choose from, n, and the objects you choose, r. The function is n!/(r!*(n-r)!). The reason the function looks like this is because you can imagine choosing r objects from a pool of n as this procedure: Sort the n objects, which can be done in n! ways. You then draw a line to separate the object you choose from the ones you don't choose. You are not interested in the order of any of the two subsets, so you divde by the amount of ways to sort each of these subsets, which are r! and (n-r)! respectively.
The 1s in the triangle can be shown to fit because nC0 and nCn are both equal to 1 as per the formula.
Now, if pascals triangle is composed of binomial coefficients, then the coefficient in row n and position n, n!/(r!*(n-r)! should be equal to the sum of the coefficients above it, which are both at row n-1, one is at position r and one is at position r-1. So if the statement is true then n!/(r!*(n-r)!) = (n-1)!/(r!*(n-1-r)!) + (n-1)!/((r-1)!*(n-r)!). We need some restrictions on n and r. We are only concerned about the numbers that are not 1s anyway, so n > 1 and 0 < r < n. These restrictions make certain multiplications legal, because they would otherwise multiply by 0. [insert maths operations] and since the sum can be show to be equal to the left hand side in the above equation, we can say that the statement is fulfilled.
Excellent, well explained :) More please!
I need to book some more time with that whiteboard!
is it also linked to, let's say, last coefficients of squares from 1 to 9 being symmetrical like Pascal's triangle? By the way great lesson and fun as always.
Hey Matt, I was reading up on Nuclear Magic Numbers. The wiki says that they are derived from binomial co-efficients but I don't see how their formula produces the numbers. Maybe there is an interesting video in that.
2, 8, 20, 28, 50, 82, and 126
I needed that for my equation, Thanks
Is there a reason why the (get ready for some terrible terminology) ending parts of the equation's exponents always equal the original power it was brought to?
For example, in your (a+b)^4 example, you get a^4, a^3b, a^2b^2, ab^3, and b^4. Each exponent, when added together, makes 4 (4=4; 3+1=4; 2+2=4; 1+3=4; 4=4).
It doesn't quite make sense to me...
I type this all out, then I realize why "sigh". It's because of the rule of when to add exponents, namely during multiplication. By virtue of the fact that we are multiplying 4 times per step (I guess is the right word?), we will get 4 items multiplied. Little derp.
+Graham Rich because you have four numbers multiplied together, each number has exponent 1 and if you add the exponents of four numbers you of course get 4. if you have (a+b)^n you would get something like (a+b)(a+b)(a+b)......(a+b) and in totalt there would be n parentheses. When you choose one number from each you effectively choose n numbers, all with exponent 1, and when you add 1 n times the result is n.
wtficecream123 Yeah, as I said, I remembered about the rule of multiplication of exponents at the very end of the comment.
I didn't catch that part... Sorry hehe
wtficecream123 Don't worry about it.
Can you explain why the numbers are the same a Pascal's Triangle?
1 C n = 1
2 C n = 1 1
3 C n = 1 2 1
And so on.
Notice how 1 4 6 4 1 is the 4th row of Pascals triangle and it appeared when you choose out of 4.
I love these videos.
I love your comments.
Favorite moment: 7:00 - 7:30 .
+standupmaths I don't know if you're gonna get this, but if you do. How do I calculate these? Looking at a task where they ask how many ways you can write 6 digits in ascending order from left to right. To do so you just write the numbers 1-9 and cross out 3 of them which can be done in 9 choose 3 ways, and then I have to calculate (a+b+c)^9 , right?!? And just how on EARTH am I going to do that?!?!
+standupmaths 6 UNIQUE numbers*
+standupmaths Digits*
I'm so amazed and unable to understand how 3 people could possibly rate this video thumbs down.
1:20 "numbered A to D"? -_-
+Sri Charan
Don't question maths!
+lolzomgz1337 oh well
I he went up to e, it wouldn't be so bad.
Sri Charan Hexadecimal?
I noticed the same thing in the main group of elevators at my office -- elevator D has "Elevator No. D" printed on the inside, next to the door (and likewise for the other elevators).
Hey Matt I have a question you might be able to answer in your next Matt Explains (BTW loving this series!)
So we doing logarithms in class and there is this basic rule saying that
a^x=b LOG(base a) b = x. So I wondered that LOG(base -2) -8 = 3 So I put it into my calculator and to my surprise I got an answer with an imaginary number. I asked my teacher and she said that it was because logs are based on a table and negative numbers are not on a table. That answer did not and still does not satisfy me. So could you please explain it.
Many thanks
Maybe, what she meant is, that your calculator works with tables, which have the logs in them, because it is an easier way to figure them out. (I don't know whether this is true). Maybe she just didn't want to say, that she didn't know the answer right ahead. People get embarrassed of these kinds of things.
I don't know, how far you have come in your maths class, maybe the whole try of an explanation following in my next comment is way to far reaching. I try my best to keep it easy, please tell me, if you did not understand it and what of it.
As far as I know (math student, 3rd semester, so you could basically say I'm making wild guesses with the help of wikipedia),
the logarithm can't be defined with negative real numbers as *base.* (Though it is possible to find things like log(base 2) of -4 )
You will always find an exception and things you can't display with real numbers. What's the logarithm of -4 base -2?
Btw, what was the imaginary number, your calculator gave you? Have you tried multiplying it? Just do a few experiments yourself.
I will try to give you a few examples and hints, please notice that I cannot answer your question easily...
The logarithm is, as you mentioned, defined as the reversal of raising to a power. Maybe you have already learnt the rules for logarithms and the natural logarithm? Then you know, that
log(base a)b = log(base c)b / log(base c)a
e.g.
log(base 2)16 = log(base 4)16 / log(base 4)2
or 4 = 2 / (1/2)
Ah yes... you can do partial exponents, too. It's similar to the root of something. _(Square root(2) = 2 ^ (1/2) )_ In fact you can have all real numbers as exponents and define a power function for it. (Do you already have functions? It's when you assign a value to a variable, following a certain rule. E.g. y = 2 * x ; I hope you do...)
So you can figure out things like log(base 2)3 and get real numbers.
Let's use y= 4^x and y=(-4)^x as an example.
log(base 4)16 = 2
log(base -4)16 would logically be 2, too.
What about this?
log(base 4)2 = 1/2 , because square root (4) = 4^(1/2) = 2
log(base -4)2 = ? What is the square root of -4? Well sqr(-4) = sqr(-1 * 4) = 2 * sqr(-1) = *2 * i*
_You get an imaginary number._
As soon as you start to insert negative values into your exponential funktion, you will get imaginary numbers for every partial exponent. Take a look at the graph at the top of this website:
www.wolframalpha.com/input/?i=%28-2%29^x
(Wolfram Alpha or similar sites are very useful, once you get into the more difficult parts of maths, they visualise things quickly)
I hope I could help you at least a little and did not just confuse you with the many concepts you have not yet been taught.
Stay curious!
Lia
Thank you very much for the answer but I am still not extremely clear on how
log(base -2) -8= 1.09 -0.42i (this is a rounded value from my TI calculator)
Video on how this relates to De Moivre's Theorem? :) interesting as always
great explanation
That was really cool!
I got a question that I would like you to explain. First since a and b in the binomial are arbitrary that would mean that the coefficients are arbitrary, meaning there are the same number of ways to choose 3 a's and 1 b than 1 a and 3 b's. Why does this symmetry occur.
Thanks Matt!
You're welcome!
to not confuse things I've learned to use subscript 4, massive C and then subscript 2
That's also correct, in fact, that is how you write it on your calculator :)
+qwertyuiopzxcfgh. Yeah, I first saw it on my calculator as nCr and was using it for puzzles, (like counting presents in the 12 Days Of Christmas song, or generalising 'triangular numbers' up to 'tetrahedral numbers' etc), long before it was introduced in school, and I just thought it stupid how they wrote it as a 2 vector.
+Dan Dart That is the more American notation. Matt is using the traditional European notation.
+Mike Djali I believe I've seen it both in subscript mostly - I first saw it in murderous maths
Yes, I'm familiar with the C notation. I actually like the ( ) notation because it gives the function an identity not directly linked to choosing (just one of its many uses).
Love your content
Also I hate expanding two lots of brackets
FOIL method, I hate it
this video teaches me lots more than my teacher did in 2+ hours
Can you systematically derive the binomial theorem from something that is so true and obvious as to be trivial? I understand how the formula works, but I don't understand how we got there in the first place. Also, I don't understand how we can be sure that the formula holds for all possible values of n.
Great video!
I already watched a 5 minute video about pascals triangle on TED-Ed, but I still won't be able to say "why" it does that. It just does.
I'm working up to the challenge of a good "why" video. Too many mathematics videos are content with pointing something amazing out but then never explaining it.
I know this is quick a bit of time later but is there a connection between the binomial theorem and homogeneous linear differential equations?
Awesome video
pls give us time travel so we can keep travelling one month into the future.(to get more of these!)
Really interesting!